Question

If \(A=\begin{bmatrix}     1 & -2 & 1 \\     2 & 1 & 3 \\ \end{bmatrix}\ B=\begin{bmatrix}     2 & 1 \\     3 & 2 \\     1 & 1 \\ \end{bmatrix}\) then (AB)' is equal to

• A. \(\begin{bmatrix}   -3 & -2 \\     10 & 7 \\ \end{bmatrix}\)
• B. \(\begin{bmatrix}   -3 & 10 \\     -2 & 7 \\ \end{bmatrix}\)
• C. \(\begin{bmatrix}   -3 &7 \\     10 & 2 \\ \end{bmatrix}\)
• D. \(\begin{bmatrix}   -3 &7 \\     10 & -2 \\ \end{bmatrix}\)

Answer 1

The Correct Option is B

First, let's calculate the product \( AB \): \[ A = \begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{pmatrix} \] The product of two matrices \( AB \) is calculated as follows: \[ AB = \begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{pmatrix} \] Now, we multiply the rows of \( A \) with the columns of \( B \): - First row, first column: \( 1 \times 2 + (-2) \times 3 + 1 \times 1 = 2 - 6 + 1 = -3 \)
First row, second column: \( 1 \times 1 + (-2) \times 2 + 1 \times 1 = 1 - 4 + 1 = -2 \)
Second row, first column: \( 2 \times 2 + 1 \times 3 + 3 \times 1 = 4 + 3 + 3 = 10 \)
Second row, second column: \( 2 \times 1 + 1 \times 2 + 3 \times 1 = 2 + 2 + 3 = 7 \)
Therefore, \( AB = \begin{pmatrix} -3 & -2 \\ 10 & 7 \end{pmatrix} \).
Now, we need to find \( (AB)^\prime \), which is the transpose of \( AB \).
The transpose of a matrix is obtained by switching its rows and columns: \[ (AB)^\prime = \begin{pmatrix} -3 & 10 \\ -2 & 7 \end{pmatrix} \]

Thus, the correct answer is \( \begin{pmatrix} -3 & 10 \\ -2 & 7 \end{pmatrix} \), which corresponds to option (B).

Answer 2

Given matrices \(A=\begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix}\) and \(B=\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}\).

We need to find \((AB)'\).

First, let's calculate \(AB\):

\(AB = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} (1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1) \end{bmatrix}\)

\(AB = \begin{bmatrix} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}\)

Now, let's find the transpose of \(AB\), which is \((AB)'\):

\((AB)' = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}' = \begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}\)

Therefore, \((AB)' = \begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}\).